In deriving the above we used \(\frac{\partial z^{\prime }}{\partial z}=\frac{x(x-1)}{(z-x)^{2}}\) . It is also to be noted that \(\lim _{z^{\prime }\rightarrow 1-x}=\lim _{z\rightarrow \infty }\) . Assuming identity block dominance Using Tauberian theorems for sufficiently heavy \(\Delta _{\mathcal {O}}\)  one can make the identity dominance [1]} assumption a bit more precise., one thereby has [2]},
\(\langle \chi (0)\chi (\infty )\mathcal {O}(1)\mathcal {O}(1-x)\rangle =x^{2(h_{\chi }-h_{\mathcal {O}})}\left(\frac{1 - (1-x)^{\alpha _{\mathcal {O}}} }{\alpha _{\mathcal {O}}} \right)^{-2h_{\chi }} ( 1-x)^{- h_{\chi }(1-\alpha _{\mathcal {O}})}, \) 
